Question: Factor completely. $-3x^2+6x+9=$
First, let's factor out the greatest common factor. If the leading coefficient is negative, we'll also factor out $-1$. The result will be something like this, where the blanks are coefficients: $_{\llcorner\!\lrcorner}\!(_{\llcorner\!\lrcorner}\! x^2+\, _{\llcorner\!\lrcorner}\! x+\, _{\llcorner\!\lrcorner}\!)$ Then, we can try to factor the sum like this: $_{\llcorner\!\lrcorner}\!(_{\llcorner\!\lrcorner}\! x+\, _{\llcorner\!\lrcorner}\!) (_{\llcorner\!\lrcorner}\! x+\, _{\llcorner\!\lrcorner}\!)$ Factor out a the greatest common factor The greatest common factor is $3$. Since the leading coefficient is negative, we'll factor out $-3$. $\begin{aligned} &\phantom{=}-3x^2+6x+9 \\\\ &=-3(x^2-2x-3) \end{aligned}$ [How did we find the greatest common factor?] Factor $x^2-2x-3$ $\begin{aligned} &\phantom{=}-3x^2+6x+9 \\\\ &=-3(x^2-2x-3) \\\\ &=-3(x+1)(x-3) \end{aligned}$ [I want to see this step in more detail.] Answer $\begin{aligned} &\phantom{=}-3x^2+6x+9 \\\\ &=-3(x+1)(x-3) \end{aligned}$ [I factored out a positive 3 instead of a negative 3. Is that wrong?]